Solving 2 Physics Questions | 65 Characters

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In summary, the conversation discussed two physics questions related to centripetal acceleration and frictional force. The first question involved a kitchen gadget for drying lettuce leaves and calculating the magnitude of the centripetal acceleration at the outer wall of the container. The second question involved a car safely negotiating a circular turn and determining the reduced speed needed to safely continue around the curve if the maximum static frictional force is reduced by a factor of three. The conversation also included calculations and hints for finding the answers to these questions.
  • #1
lotusbloom
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2 physics Questions...

1. There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 10.2 cm. When the cylinder is rotating at 2.61 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

2.A car is safely negotiating an unbanked circular turn at a speed of 19.2 m/s. The maximum static frictional force acts on the tires. Suddenly a wet patch in the road reduces the maximum static frictional force by a factor of three. If the car is to continue safely around the curve, to what speed must the driver slow the car?

please help..thanks...
 
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  • #2
1. Let's get rid of all the excess information (dear God, there is a lot of it!) - we know that the raduis is 0.102 meters and the frequency of the motion is 2.61Hz. Is this enough to find the centripetal acceleration? Let's see. You should know that ac = v2/r. But since v = ωr we can say that ac = ω2r. We have r, so we just need to find ω now... can you do it yourself? :smile:
 
  • #3
2. For the car "to continue safely around the curve", the static frictional force must match the centripetal force that the car should experience as it goes around the curve. So:

[tex]f_{s_{max}} = F_c[/tex]

Let's expand it a bit:

[tex]mg\mu _s = ma_c = m\frac{v^2}{r}[/tex]

As the mass cancels we are left with:

[tex]gr\mu _s = v^2[/tex]

So if [itex]\mu _s[/itex] is reduced by a factor of three, what must be the new speed of the car?
 
  • #4
Chen said:
1. Let's get rid of all the excess information (dear God, there is a lot of it!) - we know that the raduis is 0.102 meters and the frequency of the motion is 2.61Hz. Is this enough to find the centripetal acceleration? Let's see. You should know that ac = v2/r. But since v = ωr we can say that ac = ω2r. We have r, so we just need to find ω now... can you do it yourself? :smile:


so ac = ω2r to find W...is ac = 2.61Hz and r =0.102m put it into find W then put w in v=wr equation to find V?
 
  • #5
Chen said:
2. For the car "to continue safely around the curve", the static frictional force must match the centripetal force that the car should experience as it goes around the curve. So:

[tex]f_{s_{max}} = F_c[/tex]

Let's expand it a bit:

[tex]mg\mu _s = ma_c = m\frac{v^2}{r}[/tex]

As the mass cancels we are left with:

[tex]gr\mu _s = v^2[/tex]

So if [itex]\mu _s[/itex] is reduced by a factor of three, what must be the new speed of the car?

So the new speed will reduces by three too..is that right?
 
  • #6
lotusbloom said:
so ac = ω2r to find W...is ac = 2.61Hz and r =0.102m put it into find W then put w in v=wr equation to find V?
No, it is the frequency, f = 2.61Hz. You still don't know ac, you need to find it. ω is the angular velocity, right? So if the cylinder is rotating at 2.61 revolutions per second, how many radians does it rotate by every second? (Hint: Every circle has 2π radians.)
 
  • #7
lotusbloom said:
So the new speed will reduces by three too..is that right?
No.

[tex]gr\mu _s = v^2[/tex]

If [itex]gr\mu _s[/itex] is reduced by a factor of 3, how much is v2 reduced by? How much is v reduced by then?
 
  • #8
Chen said:
No, it is the frequency, f = 2.61Hz. You still don't know ac, you need to find it. ω is the angular velocity, right? So if the cylinder is rotating at 2.61 revolutions per second, how many radians does it rotate by every second? (Hint: Every circle has 2π radians.)

ok i think it got it...so w=5.22, r=10.2cm ( do i need to change that into m)...so then i can find ac right? then using ac = v^2/r...i can find the velocity right?
 
  • #9
Chen said:
No.

[tex]gr\mu _s = v^2[/tex]

If [itex]gr\mu _s[/itex] is reduced by a factor of 3, how much is v2 reduced by? How much is v reduced by then?

ok for this one i got an answer of 10.2 m/s...but is wrong...and i don't know why...
 
  • #10
lotusbloom said:
ok i think it got it...so w=5.22, r=10.2cm ( do i need to change that into m)...so then i can find ac right? then using ac = v^2/r...i can find the velocity right?
No, ω = 5.22π = 16.4 rad/s. And yes, you need to convert r to meters to get the units of the acceleration correct.
 
  • #11
lotusbloom said:
ok for this one i got an answer of 10.2 m/s...but is wrong...and i don't know why...
How can you get a value for v? The question asked to find how much it would be reduced by if the coefficient of friction was reduced by a factor of 3. If you reduce the value of x2 by a factor of 3, how much is x itself reduced by?
 

1. How do I solve physics questions efficiently?

There is no one-size-fits-all approach for solving physics questions, but some helpful tips include understanding the concepts and equations involved, practicing with different types of problems, and breaking down the question into smaller parts to make it more manageable.

2. How do I know which equation to use?

It is important to have a strong understanding of the concepts behind the equations, as well as knowing the variables and units involved. You can also use the given information in the question to determine which equation is most appropriate.

3. What should I do if I get stuck on a physics question?

If you are stuck on a question, take a step back and review the concepts involved. You can also try approaching the problem from a different angle or consulting a classmate or teacher for help.

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Practice is key to improving problem-solving skills in physics. Additionally, reviewing and understanding the concepts and equations involved, as well as seeking help when needed, can also aid in improving your skills.

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